Question

How do you calculate the energy of a photon of electromagnetic radiation?

Answer 1

Hint We know the relation between the frequency, wavelength of the radiation and velocity of the light as,$\text{c= }\!\!\upsilon\!\!\text{ }\!\!\lambda\!\!\text{ }$, where c is the velocity of light and is the frequency and is the wavelength.
Planck’s equation gives us the energy of electromagnetic radiation.

Complete step by step solution:
In the question it is asked how we will calculate the energy of a photon of electromagnetic radiation, we may have heard these terms earlier while dealing with quantum mechanics in physical chemistry and let’s brush up our memory quickly.
 -Electromagnetic radiations: Electromagnetic radiations are those radiations which transmits or radiates the energy in space through the periodic oscillation of the magnetic field and electric field are known as electromagnetic radiations. Or simply we can say that the electromagnetic radiations are those waves which possess electric and magnetic fields and which is perpendicular to each other and transmits energy in the space as the function of the fields. Electromagnetic radiation ranges from gamma rays to radio rays.
 -Photons-these are small units or packets of energy which possess the dual nature, the wave nature and the particle nature. So now let’s move on the solution part of the question.
We can find the values of the energy of the photon by two methods.
 First one is using Planck's equation which gives that the energy of the photons is equal to the product of Planck’s constant and frequency of the radiation.
The value of Planck’s constant is $\text{h=6}\text{.626 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-34}}}\text{Js}$
So let’s write the equation for energy of photons, $\text{E=h }\!\!\upsilon\!\!\text{ }$
If the frequency is given in Hertz then we could find the value using this equation.
If the frequency is not given and wavelength is given then we have to alter the equation to get the energy of the radiation.
We know the relation of velocity of light, frequency and wavelength as,
$\text{c= }\!\!\upsilon\!\!\text{ }\!\!\lambda\!\!\text{ }$
Alter this equation and write for frequency,
$\text{ }\!\!\upsilon\!\!\text{ =}\dfrac{\text{c}}{\text{ }\!\!\lambda\!\!\text{ }}$
Now substitute in energy equation and we get the equation as,
$E=h\dfrac{\text{c}}{\text{ }\!\!\lambda\!\!\text{ }}$
The value of c, the velocity of light is taken as $c=3\times {{10}^{8}}m/s$

Note: The value of h is different in different unit systems, hence we have to give the value of h accordingly in the equation. The value of h in the SI unit system is $\text{h=6}\text{.626 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-34}}}\text{Js}$
In the MKS system i.e. meter-kilogram-second system, the value of h is,$\text{h=4}\text{.1356 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-15}}}\text{eV}\text{.s}$
 Wavelength of the radiation will be generally in nanometers.